Find the value of $x$ for which $ff=gf$.

Question

Functions $f$ and $g$ are defined by $f:x \mapsto \frac{1}{2x+1}$, $x \neq \frac{-1}{2}$ and $g:x \mapsto x+1$. Find the value of $x$ for which $ff=gf$.

So I started in this way:

$f[f(x)]=g[f(x)]$

$f(\frac{1}{2x+1}) = g(\frac{1}{2x+1})$

$\frac{1}{2(\frac{1}{2x+1})+1}=\frac{1}{2x+1}+1$

Then I solve it in simple algebra. Is this right?

I get the wrong answer, my book says $\frac{-5}{6}$ but I get $\frac{-11}{2}$

Helppp

Answer 1

$\frac{1}{2(\frac{1}{2x+1})+1}=\frac{1}{2x+1}+1 \Leftrightarrow$

$\frac{2x+1}{2x+3}=\frac{2x+2}{2x+1}\Leftrightarrow$

$(2x+1)^2=(2x+2)(2x+3) \Leftrightarrow$

$4x^2+4x+1=4x^2+10x+6 \Leftrightarrow$

$6x+5=0$. So $x=-\frac{5}{6}$. Your book is right.

Answer 2

When I solve exactly what you wrote on your last line algebraically, I get -5/6. I suspect you made a simple mistake.